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Chapter 11. Poisson processes.

Manual for SOA Exam MLC.Chapter 11. Poisson processes.Section 11.3. Interarrival times.

c©2008. Miguel A. Arcones. All rights reserved.

Extract from:”Arcones’ Manual for SOA Exam MLC. Fall 2009 Edition”,

available at http://www.actexmadriver.com/

c©2008. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

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Chapter 11. Poisson processes. Section 11.3. Interarrival times.

Arrival time

For n ≥ 1, let Sn be the arrival time of the n–th event, i.e.

Sn = inf{t ≥ 0 : N(t) = n}.

Notice that N(Sn) = n and N(t) < n, for t < Sn.


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An useful relation between N(t) and Sn is

{Sn ≤ t}={the n−th occurrence happens before time t}={there are n or more occurrences in the interval [0, t]}={N(t) ≥ n}.

Since {Sn ≤ t} = {N(t) ≥ n},

{Sn > t} = {N(t) < n}

and

{N(t) = n} = {N(t) ≥ n} ∩ {N(t) < n + 1}={Sn ≤ t} ∩ {Sn+1 > t} = {Sn ≤ t < Sn+1}.


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Theorem 1For each integer n ≥ 1 and each t ≥ 0,

P{Gamma(n, 1) > t} = P{Poisson(t) ≤ n − 1}.

Proof.Using that

∫xn

n! e−x dx = −e−x

∑nj=0

x j

j! + c ,

P{Gamma(n, 1) > t} =∫ ∞

t

xn

n!e−x dx = −e−x

n∑j=0

x j

j!

∣∣∣∣∞t

=− e−tn∑

j=0

t j

j!= P{Poiss(t) ≤ n − 1}.


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Theorem 2Sn has a gamma distribution with parameters α = n and β =

1λ .

Proof.By Theorem 1,

P{Sn > t} = P{N(t) < n} = P{Poiss(λt) ≤ n − 1}

=P{Gamma(n, 1) > λt} = P{

Gamma(

n,1

λ

)> t

}.


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Since Sn has a gamma distribution with parameters α = n andβ = 1λ , Sn has density function

fSn(t) =λntn−1e−λt

(n − 1)!, t ≥ 0,

E [Sn] =nλ and Var(Sn) =

nλ2

.The c.d.f. of Sn is

FSn(t) = P{Sn ≤ t} = P{N(t) ≥ n} =∑∞

j=n e−λt (λt)j

j! .


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Example 1

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. Calculate:(i) P{S3 > 5}.(ii) The density of S3.(iii) Find the expected value and the variance of S3.(iv) P{S2 > 3,S5 > 7}.

Solution:


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Example 1


Solution:(i)

P{S3 > 5} = P{N(5) ≤ 2} = e−15(

1 + 15 +152

2

)=0.00003930844818.


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Example 1


Solution:(ii) S3 has a gamma distribution with α = 3 and β =

13 . Hence, the

density of S3 is

fS3(t) =33t2e−3t

3!=

9t2e−3t

2, t ≥ 0.


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Example 1


Solution:(iii)

E [S3] = 3(1/3) = 1 andVar(S3) = 3(1/3)2 = 1/3.


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(iv)

P{S2 > 3,S5 > 7} = P{N(3) ≤ 1,N(7) ≤ 4}=P{N(3) = 0,N(7) ≤ 4}+ P{N(3) = 1,N(7) ≤ 4}=P{N(3) = 0}P{N(7)− N(3) ≤ 4}

+ P{N(3) = 1}P{N(7)− N(3) ≤ 3}

=e−9e−12(

1 + 12 +122

2+

123

6+

124

24

)+ e−9(9)e−12

(1 + 12 +

122

2+

123

6

)=e−21(1237) + e−21(3357) = (3.483428261)10−6.


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Theorem 3The joint density of (S1, . . . ,Sn) is

fS1,...,Sn(s1, . . . , sn) = λne−λsn , if 0 < s1 < s2 < · · · < sn.

Proof: Given 0 < s1 < s2 < · · · < sn, take h > 0 small enough.Then,

P{S1 ∈ (s1, s1 + h], . . . ,Sn ∈ (sn, sn + h]}=P{N(s1) = 0,N(s1 + h)− N(s1) = 1,N(s2)− N(s1 + h) = 0, . . . ,

N(sn)− N(sn−1 + h) = 0,N(sn + h)− N(sn) = 1}=e−λs1e−λhλhe−λ(s2−s1−h) · · · e−λ(sn−sn−1−h)e−λhλh=λnhne−λsn

fS1,...,Sn(s1, . . . , sn)

= limh→0+

P{S1 ∈ (s1, s1 + h], . . . ,Sn ∈ (sn, sn + h]}hn

= λne−λsn .


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The distribution of (S1, . . . ,Sn−1) given Sn is uniform on0 < s1 < s2 < · · · < sn, i.e.

fS1,...,Sn−1|Sn(s1, . . . , sn−1|sn) =fS1,...,Sn(s1, . . . , sn)

fSn(sn)

=λne−λsn

λnsn−1n e−λsn(n−1)!

=(n − 1)!

sn−1n, for 0 < s1 < s2 < · · · < sn.

Since a Poisson process is a stationary process, we should expectthis distribution.


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Let Tn = Sn − Sn−1 be the time elapsed between the (n − 1)–thand the n–th event. Tn is called the interarrival between the(n − 1)–th and the n–th event.

Theorem 4{Tn}∞n=1, are independent identically distributed exponentialrandom variables having mean 1λ .

Theorem 4 says that if the rate of events is λ events per unit oftime, then the expected waiting time between events is 1λ .Theorem 4 implies that E [Tn] =

1λ , Var(Tn) =

1λ2

and Tn hasdensity function

fTn(t) = λe−λt , t ≥ 0.

Theorem 4 implies that for k1 < k2 < · · · < km,

Sk1 ,Sk2 − Sk1 ,Skm − Skm−1

are independent r.v.’s.


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Proof.We have that (S1, . . . ,Sn) = (T1,T1 + T2, . . . ,T1 + · · ·+ Tn).This transformation has Jacobian one. Hence, the density of(T1, . . . ,Tn) is

fT1,...,Tn(t1, . . . , tn) = fS1,...,Sn(t1, t1 + t2, . . . , t1 + · · ·+ tn)

=λne−λ(t1+···+tn) =n∏

j=1

λe−λtj .


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Given a sequence of independent identically distributed r.v.’s{Tn}∞n=1 with an exponential distribution with mean 1λ , define

N(t) = sup{n ≥ 0 : Sn ≤ t}, t ≥ 0,

where Sn =∑n

j=1 Tj , n ≥ 1, Sn = 0. It is possible to prove that{N(t) : t ≥ 0} is a Poisson process with rate λ. Hence, from aPoisson process we can obtain a sequence of independentidentically distributed r.v.’s with an exponential distribution, andreciprocally from a sequence of independent identically distributedr.v.’s with an exponential distribution, we can obtain a Poissonprocess.


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Example 2

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) The density of T6.(ii) Find the expected value and the variance of T6.(iii) Find Cov(T3,T8).(iv) Find Cov(S2,S9).

Solution:


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Example 2

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) The density of T6.(ii) Find the expected value and the variance of T6.(iii) Find Cov(T3,T8).(iv) Find Cov(S2,S9).Solution:(i) T6 has an exponential distribution with mean

13 . Hence, the

density of T6 isfT6(t) = 3e

−3t , t ≥ 0.


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Example 2

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) The density of T6.(ii) Find the expected value and the variance of T6.(iii) Find Cov(T3,T8).(iv) Find Cov(S2,S9).Solution:(ii)

E [T6] = (1/3),Var(T6) = (1/3)2 = 1/9.


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Example 2

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) The density of T6.(ii) Find the expected value and the variance of T6.(iii) Find Cov(T3,T8).(iv) Find Cov(S2,S9).Solution:(iii) Cov(T3,T8) = 0.


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Example 2

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) The density of T6.(ii) Find the expected value and the variance of T6.(iii) Find Cov(T3,T8).(iv) Find Cov(S2,S9).Solution:(iv)

Cov(S2,S9) = Cov(S2,S2+S9−S2) = Cov(S2,S2) = 2(1/3)2 = 2/9.


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Theorem 5Given 0 ≤ n < m and t > 0,

P{Sm − t > s|N(t) = n} = P{Sm−n > s}, s > 0,

andE [Sm|N(t) = n] = t + E [Sn−m].


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P{Sm − t > s|N(t) = n} = P{Sm−n > s}, s > 0,


Proof: We have that

P{Sm − t > s|N(t) = n} = P{Sm > t + s|N(s) = n}=P{N(t + s) < m|N(t) = n}=P{N(t + s)− N(t) < m − n|N(t) = n}=P{N(t + s)− N(t) < m − n} = P{N(s) < m − n} = P{Sm−n > s}.


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P{Sm − t > s|N(t) = n} = P{Sm−n > s}, s > 0,


Proof:

E [Sm|N(t) = n] = t + E [Sm − t|N(t) = n]

=t +

∫ ∞0

P{Sm − t > s|N(t) = n} ds

=t +

∫ ∞0

P{Sm−n > s} ds = t + E [Sm−n].


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P{Sm − t > s|N(t) = n} = P{Sm−n > s}, s > 0,


A Poisson process is a Markov process. it starts anew. The numberof occurrences observed after time t has the same distribution asthe number of occurrences observed after time zero. Hence, givenN(t) = n the waiting time after time t until the m–th occurrence isobserved has the distribution of the waiting time until the (m−n)–thoccurrence is observed.


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Example 3

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) E [S3|N(4) = 1].(ii) P{S3 > 7|N(4) = 1}.(iii) P{T3 > 7|N(4) = 1}.

Solution:


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Example 3

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) E [S3|N(4) = 1].(ii) P{S3 > 7|N(4) = 1}.(iii) P{T3 > 7|N(4) = 1}.Solution:(i) E [S3|N(4) = 1) = 4 + E [S3−1] = 4 + (3− 1)13 =

143 .


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Example 3

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) E [S3|N(4) = 1].(ii) P{S3 > 7|N(4) = 1}.(iii) P{T3 > 7|N(4) = 1}.Solution:(ii)

P{S3 > 7|N(4) = 1} = P{S3−1 > 7− 4} = P{S2 > 3}=P{N(3) ≤ 1} = e−9(1 + 9) = 0.001234.


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Example 3

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) E [S3|N(4) = 1].(ii) P{S3 > 7|N(4) = 1}.(iii) P{T3 > 7|N(4) = 1}.Solution:(ii) Alternatively,

P{S3 > 7|N(4) = 1} = P{N(7) ≤ 2|N(4) = 1} = P{N(3) ≤ 1}=e−9(1 + 9) = 0.001234.


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Example 3

Let {N(t) : t ≥ 0} be a Poisson process with rate λ = 3. Let Sndenote the time of the occurrence of the n–th event. LetTn = Sn − Sn−1 be the elapsed time between the (n − 1)–th andthe n–th event. Calculate:(i) E [S3|N(4) = 1].(ii) P{S3 > 7|N(4) = 1}.(iii) P{T3 > 7|N(4) = 1}.Solution:(iii)

P{T3 > 5|N(4) = 1} = P{T3 > 5|T1 ≤ 4 < T1 + T2}=P{T3 > 5} = e−15.


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Theorem 6Given that N(t) = n, the n arrival times S1, . . . ,Sn haveconditional density

fS1,...,Sn|N(t)=n(s1, . . . , sn) =n!

tn, 0 < s1 < s2 < · · · < sn ≤ t.

The proof of the previous theorem is in Arcones’ manual.It follows from the previous theorem that the distribution ofT1, . . . ,Tn given N(t) = n, is

fT1,...,Tn|N(t)=n(t1, . . . , tn) =n!

tn, 0 < t1, t2, . . . , tn, t1+t2+· · ·+tn ≤ t.

It also follows from the previous theorem that the distribution ofthe n arrival times S1, . . . ,Sn given N(t) = n, is the distribution oforder statistics corresponding to n independent random variablesuniformly distributed on the interval (0, t).


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Theorem 7Given s < t and k ≤ n,

P{N(s) = k|Sn = t} =(

n − 1k

) (st

)k ( t − st

)n−k−1.

Previous theorem says that given Sn = t, the number ofoccurrences in the interval [0, s] has a binomial distribution withn − 1 trials and success probability st .


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Proof:

P{N(s) = k|Sn = t}= lim

�→0+P{N(s) = k|N(t) = n,N(t − �) = n − 1}

= lim�→0+

P{N(s) = k,N(t) = n,N(t − �) = n − 1}P{N(t) = n,N(t − �) = n − 1}

= lim�→0+

e−λs (λs)k

k! e−λ(t−�−s) (λ(t−�−s))n−1−k

(n−1−k)! e−λ�λ�

e−λ(t−�) (λ(t−�))n−1

(n−1)! e−λ�λ�

=

(n − 1

k

) (st

)k ( t − st

)n−k−1.


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Example 4

Claims arrive to an insurance company website according with aPoisson rate of 100 claims per day. Suppose that in one day the10–th claim after midnight arrived at 5:15 am. Calculate theprobability that more than two claims arrived between midnightand 1:00 am.

Solution: The conditional distribution is binomial with n = 9 andp = 60(5)(60)+15 =

421 . Hence, the probability that more than two

claims arrived between midnight and 1:00 am. is

1−(

9

0

) (4

21

)0 (1721

)9−

(9

1

) (4

21

)1 (1721

)8=1− 0.1493023487− 0.3161696796 = 0.5345279717.


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Example 4

Claims arrive to an insurance company website according with aPoisson rate of 100 claims per day. Suppose that in one day the10–th claim after midnight arrived at 5:15 am. Calculate theprobability that more than two claims arrived between midnightand 1:00 am.

Solution: The conditional distribution is binomial with n = 9 andp = 60(5)(60)+15 =

421 . Hence, the probability that more than two

claims arrived between midnight and 1:00 am. is

1−(

9

0

) (4

21

)0 (1721

)9−

(9

1

) (4

21

)1 (1721

)8=1− 0.1493023487− 0.3161696796 = 0.5345279717.


Chapter 11. Poisson processes.Section 11.3. Interarrival times.

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